Here's the problem I'm having difficulties with:

Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab\,.$$

Does anyone have an idea how to do this? Any detailed solution is welcome! :)

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    $\begingroup$ What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers? $\endgroup$ Dec 8, 2018 at 18:03
  • $\begingroup$ Forgot to mention. I've edited the question. Thanks $\endgroup$ Dec 8, 2018 at 18:09
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    $\begingroup$ What are your thoughts? Put your work there and upload it. $\endgroup$
    – jayant98
    Dec 8, 2018 at 18:17
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    $\begingroup$ From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=\frac{27ab}{(a+1)(a+b)}<27\,.$$ Thus, $b\in\{1,2,\ldots,18\}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,\ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$. $\endgroup$ Dec 8, 2018 at 18:28
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    $\begingroup$ From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$. $\endgroup$
    – AdditIdent
    Dec 8, 2018 at 18:32

3 Answers 3


Using Hölder's inequality, $$27ab = (a+1)(8+b)(b+a) \geqslant \left(2\sqrt[3]{ab}+\sqrt[3]{ab} \right)^3=27ab$$

Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a \implies (a, b)=(2, 4)$.

In fact, this is the only solution among positive reals, not just positive integers.


This is a supplementary solution, where I solve for all $(a,b)\in\mathbb{Z}\times\mathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab\,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have $$(8+b)a^2+\big((8+b)(b+1)-27b\big)a+b(8+b)=0\,.$$ The discriminant of this quadratic polynomial with respect to $a$ is $$\begin{align}\big((8+b)(b+1)-27b\big)^2-4\cdot(8+b)\cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\\&=(b-4)^2(b^2-32b+4)\,.\end{align}$$ We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12\left(a^2-4a+4\right)=0\,,$$ so $a=2$. If $b\neq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$ for some integer $c$. Thus, $$d^2-c^2=252\,,$$ where $d:=b-16$.

Since $4\mid 252$ but $8\nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=\frac{d^2-c^2}{4}=63\tag{*}\,.$$ Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1)\,,\,\,(-21,-3)\,,\,\,(-9,-7)\,,\,\,(-7,-9)\,,\,\,(-3,-21)\,,\,\,(-1,-63)\,,$$ $$(1,63)\,,\,\,(3,21)\,,\,\,(7,9)\,,\,\,(9,7)\,,\,\,(21,3)\,,\text{ and }(63,1)\,.$$ Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values $$-64,-24,-16,+16,+24,+64\,.$$ Ergo, $b\in\{-48,-8,0,32,40,80\}$, resulting in the following solutions $(a,b)$: $$(80,-48)\,,\,\,(0,-8)\,,\,\,(-1,0)\,,\,\,(0,0)\,,\,\,(-5,32),(-16,40)\,,\text{ and }(-55,80)\,,$$ as well as the pair $(2,4)$ found earlier.

Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)\neq (2,4)$ take the form $$\left(-\frac{(3+r)(7+r)}{21+r},\frac{(7+r)(9+r)}{r}\right)\text{ for }r\in\mathbb{Q}\setminus\{0,-21\}\,,\tag{#}$$ and $$\left(-\frac{(9+r)(21+r)}{r(3+r)},\frac{(7+r)(9+r)}{r}\right)\text{ for }r\in\mathbb{Q}\setminus\{0,-3\}\,.\tag{@}$$ By the way, I just realized that with the transformation $r\mapsto\dfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)\neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6\pm3\sqrt{3}\text{i}$ gives rise to the pair $(a,b)=(2,4)$.)


This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.

By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,

$${8+8a\over b}+{8+b\over a}=18-a-b$$

Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.


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